# $\nabla ^2\psi$ equals $\psi -$ average value of $\psi$ at neighboring points

Let $\psi (x,y,z)$ be a scalar field. I found the following statement in Morse & Feshbach Methods of Theoretical Physics:

The limiting value of the difference between $\psi$ at a point and the average value of $\psi$ at neighboring points is $-\frac{1}{6}(dxdydz)^2\nabla ^2\psi$

By taking a small sphere centered at the point of radius $r$, I was able to show that the difference between $\psi$ at a point and the average value of $\psi$ at neighboring points equals $-\frac{1}{6}r^2\nabla ^2\psi$. But I have not been able to get the expression given in the book. Thanks.

Edit: To obtain $-\frac{1}{6}r^2\nabla ^2\psi$ I first expanded $\psi$ in a Taylor series up to quadratic terms about the point in question. Then I integrated $\psi$ (rather, its Taylor expansion up to quadratic terms) over the surface of a small sphere of radius $r$ centered at the given point and divided this by the surface are of the sphere (this gives the average value of $\psi$ in the sphere). Subtracting this average from the value of $\psi$ at the center of the sphere gives the result $-\frac{1}{6}r^2\nabla ^2\psi$.

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Would you tell us what you've done, because comparing the two expressions tells us that you're missing a factor of $4\pi$ perhaps – Approximist May 8 '11 at 19:16
@Approximist: I edited the post to explain what I had done. Why you say I could be missing a factor of $4\pi$? Thanks. – becko May 8 '11 at 19:58
@Mark@JOHA I think Morse and Feshbach mean the boundary, despite the suggestive language, after seeing their derivation of the one dimensional case for this. So the surface integration OP describes in his edit is jutified. – Approximist May 8 '11 at 23:09
I tried to write down the solution, being pretty sure that they meant a volume of a box, but your formula is certainly wrong because it doesn't even hold dimensionally. $(dx\,dy\,dz)^2$ is length to the sixth power which is surely not canceled by $\nabla^2$ so the announced "difference between $\psi$'s" doesn't have the same units as $\psi$ itself, too bad. – Luboš Motl May 9 '11 at 4:49
(This is page seven of volume one of Morse and Feshbach. Your solution is actually correct. The correction has to be obviously $$=-\frac{1}{6}(dx^2+dy^2+dz^2)\nabla^2 \psi$$. – Approximist May 9 '11 at 6:45

I think this is a typo in Morse & Feshbach Methods of Theoretical Physics. The correct expression is $-\frac{1}{6}r^2\nabla ^2\psi$ or $-\frac{1}{6}(dx^2+dy^2+dz^2)\nabla^2 \psi$.

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Formulated this way, $-\frac16\,r^2\,\nabla^2\psi$ is indeed the correct result. While it is certainly more elegant to express such things by means of differential forms rather than plain functions with finite $r$, the result as given in the book is not a meaningful differential form at all, since $$(\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z)^2 = \mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\wedge\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z$$ $$= (-1)(-1)\mathrm{d}x\wedge\mathrm{d}x\wedge\mathrm{d}y\wedge\mathrm{d}z\wedge\mathrm{d}y\wedge\mathrm{d}z = 0,$$ due to $\mathrm{d}x_i\wedge\mathrm{d}x_j=-\mathrm{d}x_j\wedge\mathrm{d}x_i$. The correct way might be something like $$-\frac16\,(\mathrm{d}y\mathrm{d}z+\mathrm{d}z\mathrm{d}x+\mathrm{d}x\mathrm{d}y)\,\nabla^2\psi$$ but I'm not sure if this would much more meaningful. The plain finite-small-$r$ expression is probably not the worst after all.

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## protected by Qmechanic♦May 21 '13 at 11:07

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